Theorem: For any integers p and q define the Abel-sum type sequence by n ----- \ k (k - 1 + p) (n - k + q) A[n](r, s) = ) binomial(n, k) x (r + k) (s - k) / ----- k = 0 Then we have the following functional recurrence expressing A[n](r.s) in ter\ ms of , A[n - 1](r, s), A[n - 2](r, s) : A[n](r, s) = (n x + r x) A[n - 1](r + 1, s - 1) + s A[n - 1](r, s) + (-n r x - n s x + r x + s x) A[n - 2](r + 1, s - 1) and in Maple notation A[n](r,s) = (n*x+r*x)*A[n-1](r+1,s-1)+s*A[n-1](r,s)+(-n*r*x-n*s*x+r*x+s*x)*A[n-\ 2](r+1,s-1) Proof: We claim that: Let a(n,k,r,s) be the summand on the sum defining A[n](r,s) in other words k (k - 1 + p) (n - k + q) a(n, k, r, s) = binomial(n, k) x (r + k) (s - k) and in Maple notation a(n,k,r,s) = binomial(n,k)*x^k*(r+k)^(k-1+p)*(s-k)^(n-k+q) Then the following identity is true x (r + s) (n + 1) a(n, k, r, s) - (n + r + 1) x a(n + 1, k, r, s) + (-1 - s) a(n + 1, k + 1, r - 1, 1 + s) + a(n + 2, k + 1, r - 1, 1 + s) = 0 and in Maple notation: x*(r+s)*(n+1)*a(n,k,r,s)-(n+r+1)*x*a(n+1,k,r,s)+(-1-s)*a(n+1,k+1,r-1,1+s)+a(n+2 ,k+1,r-1,1+s) = 0 The proof of this identity is routine (divide by a(n,k,r,s), simplify each t\ erm,and now each term is a rational function. Now add them all up and \ verify that they add up to zero.) Now sum it from k=0 to k=n, which is the same as from k=-infinity to k=infin\ ity (since it vanishes for k<0 and k>n x (r + s) (n + 1) A[n](r, s) - (n + r + 1) x A[n + 1](r, s) + (-1 - s) A[n + 1](r - 1, 1 + s) + A[n + 2](r - 1, 1 + s) = 0 replacing n by, n - 2, changing variables, and moving a[n](r,s) to the left \ side, yields the statement of the thereom. QED. ------------------------------------------------- This took, 0.124, seconds. ------------------------------------------- Theorem: For any integers p and q define the Abel-sum type sequence by n ----- \ 2 k (k - 1 + p) (n - k + q) A[n](r, s) = ) binomial(n, k) x (r + k) (s - k) / ----- k = 0 Then we have the following functional recurrence relating , A[n](r, s), A[n + 1](r, s), A[n + 2](r, s), A[n + 3](r, s) 2 3 (r + s) (n + 1) (n + 2 r - 1) (n + r - s - 3) x A[n](r, s) - ------------------------------------------------------------ - 2 (r + s) 2 (n - 2 s - 3) (s + 3) (n + r - s - 1) 2 2 2 2 2 (n + 1) (n + 2 n r - 2 n s + r - 2 r s + s - 4 n - 5 r + 3 s + 3) x / 2 A[n](r - 1, 1 + s) / ((n - 2 s - 3) (s + 3) (n + r - s - 1)) / 2 x (r + s) (n + 1) A[n](r - 2, s + 2) - ------------------------------------- 2 (s + 3) 3 2 x (n + 2 r - 1) (n + r + 1) (n + r - s - 3) A[n + 1](r, s) 4 + ------------------------------------------------------------ + (2 n 2 (n - 2 s - 3) (s + 3) (n + r - s - 1) 3 3 2 2 2 3 2 2 + 10 n r - 2 n s + 17 n r - 10 n r s + 9 n r - 17 n r s + 2 n r s 3 2 2 3 2 2 2 2 - 6 r s + 6 r s - 8 n - 28 n r + 6 n s - 24 n r + 22 n r s - 6 n s 3 2 2 2 2 2 + 5 r + 23 r s - 14 r s + 7 n + 3 n r - 21 n s - 19 r - 46 r s + 2 s 2 / - 4 n - 7 r + 7 s + 3) x A[n + 1](r - 1, 1 + s) / ((n - 2 s - 3) / 2 4 3 3 2 2 2 (s + 3) (n + r - s - 1)) + (n + 5 n r - n s + 4 n r - 14 n r s 2 2 2 2 3 2 2 3 3 - 3 n s - 10 n r s + 13 n r s + 3 n s + 6 r s - 6 r s - 6 n 2 2 2 2 2 2 - 21 n r + 13 n s - 12 n r + 40 n r s - 24 n s + 10 r s - 37 r s 3 2 2 2 + 17 s + 20 n + 25 n r - 73 n s - 2 r - 66 r s + 61 s - 42 n - 33 r / 2 + 69 s + 27) x A[n + 1](r - 2, s + 2) / ((n - 2 s - 3) (s + 3) / (n + r - s - 1)) + A[n + 1](r - 3, s + 3) - 2 (n + r + 1) 2 2 / (n + 2 n r + 2 n + 5 r - 1) (n + r - s - 3) x A[n + 2](r - 1, 1 + s) / / 2 4 3 ((n + 2) (n - 2 s - 3) (s + 3) (n + r - s - 1)) - 2 x (n + 2 n r 3 2 2 2 2 2 3 2 2 2 - 2 n s + n r - 2 n r s + n s - n + n r - 3 n s + 2 n r 2 2 2 2 - 8 n r s + 2 n s - 8 n - 15 n r + 3 n s - r - 8 r s - s - 15 r - 3 s) / 2 A[n + 2](r - 2, s + 2) / ((n + 2) (n - 2 s - 3) (s + 3) (n + r - s - 1)) / 2 2 (n - 2 n s - n - 5 s - 9) A[n + 2](r - 3, s + 3) - --------------------------------------------------- (n + 2) (s + 3) (n - 2 s - 3) x (n + 3) (n + 2 r) (n + r - s - 3) A[n + 3](r - 2, s + 2) + ---------------------------------------------------------- 2 (n + 2) (n - 2 s - 3) (s + 3) (n + r - s - 1) (n + 3) (n - 2 s - 4) A[n + 3](r - 3, s + 3) + -------------------------------------------- = 0 2 (n - 2 s - 3) (n + 2) (s + 3) and in Maple notation -(r+s)*(n+1)^2*(n+2*r-1)*(n+r-s-3)*x^3/(n-2*s-3)/(s+3)^2/(n+r-s-1)*A[n](r,s)-2* (r+s)*(n+1)^2*(n^2+2*n*r-2*n*s+r^2-2*r*s+s^2-4*n-5*r+3*s+3)*x^2/(n-2*s-3)/(s+3) ^2/(n+r-s-1)*A[n](r-1,1+s)-x*(r+s)*(n+1)^2/(s+3)^2*A[n](r-2,s+2)+x^3*(n+2*r-1)* (n+r+1)^2*(n+r-s-3)/(n-2*s-3)/(s+3)^2/(n+r-s-1)*A[n+1](r,s)+(2*n^4+10*n^3*r-2*n ^3*s+17*n^2*r^2-10*n^2*r*s+9*n*r^3-17*n*r^2*s+2*n*r*s^2-6*r^3*s+6*r^2*s^2-8*n^3 -28*n^2*r+6*n^2*s-24*n*r^2+22*n*r*s-6*n*s^2+5*r^3+23*r^2*s-14*r*s^2+7*n^2+3*n*r -21*n*s-19*r^2-46*r*s+2*s^2-4*n-7*r+7*s+3)*x^2/(n-2*s-3)/(s+3)^2/(n+r-s-1)*A[n+ 1](r-1,1+s)+(n^4+5*n^3*r-n^3*s+4*n^2*r^2-14*n^2*r*s-3*n^2*s^2-10*n*r^2*s+13*n*r *s^2+3*n*s^3+6*r^2*s^2-6*r*s^3-6*n^3-21*n^2*r+13*n^2*s-12*n*r^2+40*n*r*s-24*n*s ^2+10*r^2*s-37*r*s^2+17*s^3+20*n^2+25*n*r-73*n*s-2*r^2-66*r*s+61*s^2-42*n-33*r+ 69*s+27)*x/(n-2*s-3)/(s+3)^2/(n+r-s-1)*A[n+1](r-2,s+2)+A[n+1](r-3,s+3)-2*(n+r+1 )*(n^2+2*n*r+2*n+5*r-1)*(n+r-s-3)*x^2/(n+2)/(n-2*s-3)/(s+3)^2/(n+r-s-1)*A[n+2]( r-1,1+s)-2*x*(n^4+2*n^3*r-2*n^3*s+n^2*r^2-2*n^2*r*s+n^2*s^2-n^3+n^2*r-3*n^2*s+2 *n*r^2-8*n*r*s+2*n*s^2-8*n^2-15*n*r+3*n*s-r^2-8*r*s-s^2-15*r-3*s)/(n+2)/(n-2*s-\ 3)/(s+3)^2/(n+r-s-1)*A[n+2](r-2,s+2)-2*(n^2-2*n*s-n-5*s-9)/(n+2)/(s+3)/(n-2*s-3 )*A[n+2](r-3,s+3)+x*(n+3)*(n+2*r)*(n+r-s-3)/(n+2)/(n-2*s-3)/(s+3)^2/(n+r-s-1)*A [n+3](r-2,s+2)+(n+3)*(n-2*s-4)/(n-2*s-3)/(n+2)/(s+3)^2*A[n+3](r-3,s+3) = 0 Proof: We claim that: Let a(n,k,r,s) be the summand on the sum defining A[n](r,s) in other words 2 k (k - 1 + p) (n - k + q) a(n, k, r, s) = binomial(n, k) x (r + k) (s - k) and in Maple notation a(n,k,r,s) = binomial(n,k)^2*x^k*(r+k)^(k-1+p)*(s-k)^(n-k+q) Then the following identity is true 2 3 (r + s) (n + 1) (n + 2 r - 1) (n + r - s - 3) x a(n, k, r, s) - --------------------------------------------------------------- - 2 (r + s) 2 (n - 2 s - 3) (s + 3) (n + r - s - 1) 2 2 2 2 2 (n + 1) (n + 2 n r - 2 n s + r - 2 r s + s - 4 n - 5 r + 3 s + 3) x / 2 a(n, k + 1, r - 1, 1 + s) / ((n - 2 s - 3) (s + 3) (n + r - s - 1)) / 2 x (r + s) (n + 1) a(n, k + 2, r - 2, s + 2) - -------------------------------------------- 2 (s + 3) 3 2 x (n + 2 r - 1) (n + r + 1) (n + r - s - 3) a(n + 1, k, r, s) 4 + --------------------------------------------------------------- + (2 n 2 (n - 2 s - 3) (s + 3) (n + r - s - 1) 3 3 2 2 2 3 2 2 + 10 n r - 2 n s + 17 n r - 10 n r s + 9 n r - 17 n r s + 2 n r s 3 2 2 3 2 2 2 2 - 6 r s + 6 r s - 8 n - 28 n r + 6 n s - 24 n r + 22 n r s - 6 n s 3 2 2 2 2 2 + 5 r + 23 r s - 14 r s + 7 n + 3 n r - 21 n s - 19 r - 46 r s + 2 s 2 / - 4 n - 7 r + 7 s + 3) x a(n + 1, k + 1, r - 1, 1 + s) / ((n - 2 s - 3) / 2 4 3 3 2 2 2 (s + 3) (n + r - s - 1)) + (n + 5 n r - n s + 4 n r - 14 n r s 2 2 2 2 3 2 2 3 3 - 3 n s - 10 n r s + 13 n r s + 3 n s + 6 r s - 6 r s - 6 n 2 2 2 2 2 2 - 21 n r + 13 n s - 12 n r + 40 n r s - 24 n s + 10 r s - 37 r s 3 2 2 2 + 17 s + 20 n + 25 n r - 73 n s - 2 r - 66 r s + 61 s - 42 n - 33 r / 2 + 69 s + 27) x a(n + 1, k + 2, r - 2, s + 2) / ((n - 2 s - 3) (s + 3) / (n + r - s - 1)) + a(n + 1, k + 3, r - 3, s + 3) - 2 (n + r + 1) 2 2 (n + 2 n r + 2 n + 5 r - 1) (n + r - s - 3) x / 2 a(n + 2, k + 1, r - 1, 1 + s) / ((n + 2) (n - 2 s - 3) (s + 3) / 4 3 3 2 2 2 2 2 (n + r - s - 1)) - 2 x (n + 2 n r - 2 n s + n r - 2 n r s + n s 3 2 2 2 2 2 - n + n r - 3 n s + 2 n r - 8 n r s + 2 n s - 8 n - 15 n r + 3 n s 2 2 / - r - 8 r s - s - 15 r - 3 s) a(n + 2, k + 2, r - 2, s + 2) / ((n + 2) / 2 (n - 2 s - 3) (s + 3) (n + r - s - 1)) 2 2 (n - 2 n s - n - 5 s - 9) a(n + 2, k + 3, r - 3, s + 3) - ---------------------------------------------------------- (n + 2) (s + 3) (n - 2 s - 3) x (n + 3) (n + 2 r) (n + r - s - 3) a(n + 3, k + 2, r - 2, s + 2) + ----------------------------------------------------------------- 2 (n + 2) (n - 2 s - 3) (s + 3) (n + r - s - 1) (n + 3) (n - 2 s - 4) a(n + 3, k + 3, r - 3, s + 3) + --------------------------------------------------- = 0 2 (n - 2 s - 3) (n + 2) (s + 3) and in Maple notation: -(r+s)*(n+1)^2*(n+2*r-1)*(n+r-s-3)*x^3/(n-2*s-3)/(s+3)^2/(n+r-s-1)*a(n,k,r,s)-2 *(r+s)*(n+1)^2*(n^2+2*n*r-2*n*s+r^2-2*r*s+s^2-4*n-5*r+3*s+3)*x^2/(n-2*s-3)/(s+3 )^2/(n+r-s-1)*a(n,k+1,r-1,1+s)-x*(r+s)*(n+1)^2/(s+3)^2*a(n,k+2,r-2,s+2)+x^3*(n+ 2*r-1)*(n+r+1)^2*(n+r-s-3)/(n-2*s-3)/(s+3)^2/(n+r-s-1)*a(n+1,k,r,s)+(2*n^4+10*n ^3*r-2*n^3*s+17*n^2*r^2-10*n^2*r*s+9*n*r^3-17*n*r^2*s+2*n*r*s^2-6*r^3*s+6*r^2*s ^2-8*n^3-28*n^2*r+6*n^2*s-24*n*r^2+22*n*r*s-6*n*s^2+5*r^3+23*r^2*s-14*r*s^2+7*n ^2+3*n*r-21*n*s-19*r^2-46*r*s+2*s^2-4*n-7*r+7*s+3)*x^2/(n-2*s-3)/(s+3)^2/(n+r-s -1)*a(n+1,k+1,r-1,1+s)+(n^4+5*n^3*r-n^3*s+4*n^2*r^2-14*n^2*r*s-3*n^2*s^2-10*n*r ^2*s+13*n*r*s^2+3*n*s^3+6*r^2*s^2-6*r*s^3-6*n^3-21*n^2*r+13*n^2*s-12*n*r^2+40*n *r*s-24*n*s^2+10*r^2*s-37*r*s^2+17*s^3+20*n^2+25*n*r-73*n*s-2*r^2-66*r*s+61*s^2 -42*n-33*r+69*s+27)*x/(n-2*s-3)/(s+3)^2/(n+r-s-1)*a(n+1,k+2,r-2,s+2)+a(n+1,k+3, r-3,s+3)-2*(n+r+1)*(n^2+2*n*r+2*n+5*r-1)*(n+r-s-3)*x^2/(n+2)/(n-2*s-3)/(s+3)^2/ (n+r-s-1)*a(n+2,k+1,r-1,1+s)-2*x*(n^4+2*n^3*r-2*n^3*s+n^2*r^2-2*n^2*r*s+n^2*s^2 -n^3+n^2*r-3*n^2*s+2*n*r^2-8*n*r*s+2*n*s^2-8*n^2-15*n*r+3*n*s-r^2-8*r*s-s^2-15* r-3*s)/(n+2)/(n-2*s-3)/(s+3)^2/(n+r-s-1)*a(n+2,k+2,r-2,s+2)-2*(n^2-2*n*s-n-5*s-\ 9)/(n+2)/(s+3)/(n-2*s-3)*a(n+2,k+3,r-3,s+3)+x*(n+3)*(n+2*r)*(n+r-s-3)/(n+2)/(n-\ 2*s-3)/(s+3)^2/(n+r-s-1)*a(n+3,k+2,r-2,s+2)+(n+3)*(n-2*s-4)/(n-2*s-3)/(n+2)/(s+ 3)^2*a(n+3,k+3,r-3,s+3) = 0 The proof of this identity is routine (divide by a(n,k,r,s), simplify each t\ erm,and now each term is a rational function. Now add them all up and \ verify that they add up to zero.) Now sum it from k=0 to k=n, which is the same as from k=-infinity to k=infin\ ity (since it vanishes for k<0 and k>n). Since the above recurrence is \ free from k, we get the claimed recurrence. QED. ------------------------------------------------- This took, 43.759, seconds. ------------------------------------------- Theorem: For any integers p and q define the Abel-sum type sequence by n ----- k (k - 1 + p) (n - k + q) \ x (r + k) (s - k) A[n](r, s) = ) ---------------------------------------- / 2 ----- (k!) (n - k)! k = 0 Then we have the following functional recurrence expressing A[n](r.s) in ter\ ms of , A[n - 1](r, s), A[n - 2](r, s), A[n - 3](r, s) : x (n + r) A[n - 1](r + 1, s - 1) A[n](r, s) = -------------------------------- 2 n 2 (2 n - 2 n s - 2 n + s) s A[n - 1](r, s) + ----------------------------------------- 2 (n - 1 - s) n 2 2 (n + 2 n r - 2 r s - s - n - r) x A[n - 2](r + 1, s - 1) - ---------------------------------------------------------- 2 (n - 1 - s) n 2 (n - s) s A[n - 2](r, s) - ------------------------- 2 (n - 1 - s) n 2 (n r + n s - r s - s ) x A[n - 3](r + 1, s - 1) + ----------------------------------------------- 2 (n - 1 - s) n and in Maple notation A[n](r,s) = x*(n+r)/n^2*A[n-1](r+1,s-1)+(2*n^2-2*n*s-2*n+s)*s/(n-1-s)/n^2*A[n-1 ](r,s)-(n^2+2*n*r-2*r*s-s^2-n-r)*x/(n-1-s)/n^2*A[n-2](r+1,s-1)-(n-s)*s^2/(n-1-s )/n^2*A[n-2](r,s)+(n*r+n*s-r*s-s^2)*x/(n-1-s)/n^2*A[n-3](r+1,s-1) Proof: We claim that: Let a(n,k,r,s) be the summand on the sum defining A[n](r,s) in other words k (k - 1 + p) (n - k + q) x (r + k) (s - k) a(n, k, r, s) = ---------------------------------------- 2 (k!) (n - k)! and in Maple notation a(n,k,r,s) = 1/k!^2/(n-k)!*x^k*(r+k)^(k-1+p)*(s-k)^(n-k+q) Then the following identity is true (r + s) (n - s + 2) a(n, k, r, s) --------------------------------- (n - s + 1) (n + r + 2) 2 2 (n + 2 n r - 2 r s - s + 3 n + 3 r + 2) a(n + 1, k, r, s) - ----------------------------------------------------------- (n - s + 1) (n + r + 2) 2 (1 + s) (n - s + 2) a(n + 1, k + 1, r - 1, 1 + s) - -------------------------------------------------- + a(n + 2, k, r, s) x (n - s + 1) (n + r + 2) 2 (1 + s) (2 n - 2 n s + 8 n - 5 s + 7) a(n + 2, k + 1, r - 1, 1 + s) + -------------------------------------------------------------------- x (n - s + 1) (n + r + 2) 2 (n + 3) a(n + 3, k + 1, r - 1, 1 + s) - -------------------------------------- = 0 x (n + r + 2) and in Maple notation: (r+s)*(n-s+2)/(n-s+1)/(n+r+2)*a(n,k,r,s)-(n^2+2*n*r-2*r*s-s^2+3*n+3*r+2)/(n-s+1 )/(n+r+2)*a(n+1,k,r,s)-(1+s)^2*(n-s+2)/x/(n-s+1)/(n+r+2)*a(n+1,k+1,r-1,1+s)+a(n +2,k,r,s)+(1+s)*(2*n^2-2*n*s+8*n-5*s+7)/x/(n-s+1)/(n+r+2)*a(n+2,k+1,r-1,1+s)-(n +3)^2/x/(n+r+2)*a(n+3,k+1,r-1,1+s) = 0 The proof of this identity is routine (divide by a(n,k,r,s), simplify each t\ erm,and now each term is a rational function. Now add them all up and \ verify that they add up to zero.) Now sum it from k=0 to k=n, which is the same as from k=-infinity to k=infin\ ity (since it vanishes for k<0 and k>n (r + s) (n - s + 2) A[n](r, s) ------------------------------ (n - s + 1) (n + r + 2) 2 2 (n + 2 n r - 2 r s - s + 3 n + 3 r + 2) A[n + 1](r, s) - -------------------------------------------------------- (n - s + 1) (n + r + 2) 2 (1 + s) (n - s + 2) A[n + 1](r - 1, 1 + s) - ------------------------------------------- + A[n + 2](r, s) x (n - s + 1) (n + r + 2) 2 (1 + s) (2 n - 2 n s + 8 n - 5 s + 7) A[n + 2](r - 1, 1 + s) + ------------------------------------------------------------- x (n - s + 1) (n + r + 2) 2 (n + 3) A[n + 3](r - 1, 1 + s) - ------------------------------- = 0 x (n + r + 2) replacing n by, n - 3, changing variables, and moving a[n](r,s) to the left \ side, yields the statement of the thereom. QED. ------------------------------------------------- This took, 0.111, seconds. -------------------------------------------